How do you graph the curve whose parametric equations are given and show its orientation given #x = sqrt{t} + 4#, #y = sqrt{t} - 4#, where #t>=0#?

1 Answer
Dec 3, 2017

Pleasesee below.

Explanation:

For a parametric equation #f)(x,y)=(x(t),y(t))#, you put different values of #t#, ofcourse where #t>=0#, to get different pairs of values for #x# and #y# to get sets of points, joining which, we get the desired curve.

Here we have #x=sqrtt+4# and #y=sqrtt-4#. Let us consider #t=0,1,4,9,16,25,36# - note that we have intentionally selected square numbers, so that getting pair of values of #x# and #y# is easy.

We get #(4,-4),(5,-3),(6,-2),(7,-1),(8,0),(9,1),(10,2)# and joining them we get the following graph nand this is a striaght line, shown below. Also observe that given #x=sqrtt+4# and #y=sqrtt-4#, subtracting latter from former eliminates #t# and we get equation of line #x-y=8#. What does #sqrtt# does to this? It just restricts line to #x>=4# or #y>=-4#.

graph{x-y=8 [4, 24, -7, 3]}